to the sides ED, EF, and the angle ABC is equal to the angle

DEF.

LESSON II.

DEF. VI. If one right line, standing upon another, make the adjacent angles equal to one another, each of these angles is called a right angle, and the

right line which stands upon the other is called a perpendicular to it.

C

A D B

Thus if the right line CD stand upon the right line AB, and make the angles CDA, CDB equal to one another, each of these angles is called a right angle; and the line CD is called a perpendicular to the line AB. [See NOTE G.]

DEM. In the triangles FCE, GCE, the side cr is equal to CG, by DEF. 3.; and the side EF to EG, by construction: and the side CE is common. Hence, by ART. 6, the angles CEF, CEG are equal; that is, by DEF. VI., CE is perpendicular to AB. This, &c.

The right line is supposed capable of being produced if necessary; for the given point might be so situated with respect to it that no perpendicular could be drawn to it unless produced.

PROB. VII. To draw a perpendicular to a given right line, from a point in it.

Let AB be the given line, c the point in it. It is required to draw from c a perpendicular to AB.

CONS. At different sides of c take CD, CE, equal to each other, by PROB. III. Upon DE describe the equilateral triangle DFE, by PROB. I. and join FC. Then, Fc is perpendicular to AB.

A D

F

E B

DEM. In the triangles DFC, EFC, the side Fc is common, the sides DF, FE, are equal to each other, and the sides DC, CE, are also equal to each other. Hence, by ART. 6, the angle FCD is equal to the angle FCE; that is, Fc is perpendicular to AB, by DEF. VI.

The right line in this problem also is supposed capable of being produced if necessary; as the given point might be at one of its

extremities.

DEF. VII. An angle greater than a right angle is called an obtuse angle; and an angle less than a right angle is called an acute angle.

Thus, if CD be perpendicular to AB, the right line DE from the point D makes with AB two angles; one of which EDR, being greater than the right angle CDA, is called obtuse, and the other, EDA, being less than the right angle CDA, is called acute.

E

A

ART. 8. All right Angles are equal.

Let CDB, GHE be any two right angles. Then the angle CDB is equal to the angle GHE.

AD B

V

EHT

DEM. Produce BD through D to any point A, and since CDB is granted a right angle, the line CD must form with the line AB the two angles CDA, CDB, equal, by DEF. VI.

PROB. VIII. At a given point in a given indefinite right line, to draw a right line making, with the given one, an angle equal to a given angle.

Let BAC be the given angle, DE the given line, and the given point in it. It is required to make an angle equal to BAC at the point F, one side of which shall be the right line FE.

CONS. On AB take any point G, and from AC cut off AH equal to AG, by PROB. III.; join HG. Likewise, from the line FE cut off FI equal to AG, by PROB. III.; and with F as a centre, describe the

circle NLI, at the distance FI. From the line IE cut off IK equal to GH, by PROB. III.; and with 1 as a centre, describe the circle

to the sides ED, EF, and the angle ABC is equal to the angle

DEF.

LESSON II.

C.

DEF. VI. If one right line, standing upon another, make the adjacent angles equal to one another, each of these angles is called a right angle, and the right line which stands upon the other is called a perpendicular to it.

A

D B

Thus if the right line CD stand upon the right line AB, and make the angles CDA, CDB equal to one another, each of these angles is called a right angle; and the line CD is called a perpendicular to the line AB. [See NOTE G.]

DEM. In the triangles FCE, GCE, the side cF is equal to CG, by DEF. 3.; and the side EF to EG, by construction: and the side ce is common. Hence, by ART. 6, the angles CEF, CEG are equal; that is, by DEF. VI., CE is perpendicular to AB. This, &c. The right line is supposed capable of being produced if necessary; for the given point might be so situated with respect to it that no perpendicular could be drawn to it unless produced.

PROB. VII. To draw a perpendicular to a given right line, from a point in it.

Let AB be the given line, c the point in it. It is required to draw from c a perpendicular to AB.

CONS. At different sides of c take CD, CE, equal to each other, by PROB. III. Upon DE describe the equilateral triangle DFE, by PROB. I. and join Fc. Then, FC is perpendicular to AB.

A D

F

E B

DEM. In the triangles DFC, EFC, the side Fc is common, the sides DF, FE, are equal to each other, and the sides DC, CE, are also equal to each other. Hence, by ART. 6, the angle FCD is equal to the angle FCE; that is, Fc is perpendicular to AB, by DEF. VI.

The right line in this problem also is supposed capable of being produced if necessary; as the given point might be at one of its extremities.

DEF. VII. An angle greater than a right angle is called an obtuse angle; and an angle less than a right angle is called an acute angle.

Thus, if CD be perpendicular to AB, the right line DE from the point D makes with AB two angles; one of which EDR, being greater than the right angle CDA, is called obtuse, and the other, EDA, being less than the right angle CDA, is called acute.

E

A

ART. 8. All right Angles are equal.

Let CDB, GHE be any two right angles. Then the angle CDB is equal to the angle GHE.

ADB

B

Ꮐ

V

EHT

DEM. Produce BD through D to any point A, and since CDB is granted a right angle, the line CD must form with the line AB the two angles CDA, CDB, equal, by Def. VI.

PROB. VIII. At a given point in a given indefinite right line, to draw a right line making, with the given one, an angle equal to a given angle.

Let BAC be the given angle, DE the given line, and the given point in it. It is required to make an angle equal to BAC at the point F, one side of which shall be the right line FE.

CONS. On AB take any point G, and from Ac cut off AH equal to AG, by PROB. III.; join HG. Likewise, from the line FE cut off FI equal to AG, by PROB. III.; and with F as a centre, describe the

circle NLI, at the distance FI. From the line IE cut off IK equal to GH, by PROB. III.; and with 1 as a centre, describe the circle

For the same reason, producing EH through H to any point F, the angles GHE, GHF are equal. Now conceive the line AB so applied to the line EF, that the point D may fall on the point H, and the line AB may fall along the line EF; also that DC may be at the same side of EF with HG: then, DC would necessarily fall along HG. For suppose it to fall in any other direction, as HI; consequently, the angles GHE, GHF, being equal, the angles IHE, IHF are unequal,—that is, the angles CDA, CDB (which are the same as IHE, IHF) would be, on this supposition, unequal. But they have been proved equal; therefore this supposition is false, the line DC would not fall in any other direction but HG. Hence the angles CDB, GHE must be equal, inasmuch as the point D being on the point н, and the line AB on the line EF, the side DC would fall on the side HG *. This &c.

This principle is omitted in Euclid; but is necessary to a rigorous system of demonstration, such as the Elements of Geometry should exhibit.

OLK, at the distance IK. Join the points F and I with the point L, where these circles intersect. Then the angle IFL is equal to the angle BAC.

DEM. Because AG is equal to FI by construction, and FI is equal to FL, by DEF. 3; therefore AG and FL are equal t. But AG is equal to AH by construction; and therefore AH is likewise equal to FL +. Finally, GH is equal to IK, by construction, and IK is equal to IL, by DEF. 3: therefore GH and IL are equal+. Hence, in the two triangles GAH, IFL, since the three sides AG, AH, and GH, are respectively equal to the three sides FI, FL, and IL, the angle GAH opposite to GH in the one, is equal to the angle IFL opposite to the equal side IL in the other, by ART. 6. This, &c. [See NOTE H.]

In the common Euclids this problem is enounced thus: "At a given point in a given straight line to make a rectilineal angle equal to a given rectilineal angle." This is imperfect, the condition of the given right line forming one side of the required angle being omitted.